A2 - Ch 24a Winter Term 2009‐10 Introduction to...

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Unformatted text preview: Ch 24a Winter Term 2009‐10 Introduction to Biophysical Chemistry Problem Set # 2 Distributed: 29 January 2010 Due: 8 February 2010, 1 p.m. 1. (25 points) (a) The π→ π* transition of the amino acid tryptophan in aqueous solution at pH 5 to 7 peaks at 220 nm with an extinction coefficient ε of 3.5 x 104 and a half‐width at half height of ~ 10 nm. Assume that the absorption band can be approximated by a Gaussian in frequency space and estimate the Einstein coefficient A for spontaneous fluorescence in sec‐1. (20 points) (b) Use the result above to calculate the intrinsic fluorescence lifetime of tryptophan, namely, the radiative lifetime τR. If the quantum yield φF for tryptophan in a protein is 0.3, what will be the corresponding fluorescence decay time τF? (5 points) a) v = c / λ = 3 ⋅ 1010 /(220 ⋅ 10−7 ) = 1.36 ⋅ 1015 sec−1 to desicribe ε(v ), one needs : v a = c / λ = 3 ⋅ 1010 /(210 ⋅ 10−7 ) = 1.428 ⋅ 1015 sec−1 v b = c / λ = 3 ⋅ 1010 /(230 ⋅ 10−7 ) = 1.304 ⋅ 1015 sec−1 Full width at half maximum for a gaussian function = 2 2ln2σ = v a − v b = 0.1243 ⋅ 1015 σ = 5.266 ⋅ 1013 −( v −1.36⋅1015 )2 ε(v ) = 3.5 ⋅ 10 ⋅ e 2⋅(5.266⋅10 4 13 2 ) ε(v ) 8000πv 3 A= ∫ v dv 2 Nc log e band −(v − 1.36 ⋅ 1015 ) 2 exp ∞ 8000π (1.36 ⋅ 1015 ) 3 2 ⋅ (5.266 ⋅ 1013 ) 2 3.5 ⋅ 10 4 dv = 9.13 ⋅ 10 8 molecule/sec = ∫−∞ 23 10 2 6 ⋅ 10 ⋅ (3 ⋅ 10 ) ⋅ 0.434 v b) 1 = 1.09 ⋅ 10−9 sec A τ F = 0.3 ⋅ τ R = 3.28 ⋅ 10−10 sec τR = 2. (10 points) For bacterial reaction centers, the quantum yield of fluorescence from the excited state of the special pair, a highly interacting pair of bacteriochlorophylls, is 10‐4. The intrinsic or natural radiative decay rate of the bacteriochlororphyll dimer is 108 sec‐1. Assuming that there are no non‐radiative processes other than electron transfer, what is the rate of transfer of an electron from the excited state of the bacteriochlorophyll dimer to the bacteriochlorophyll monomer before it is transferred to the bacteriopheophytin and subsequently to the primary quinone. (Adapted from James P. Allen, Chapter 14 problem 14.17. For background, see James P. Allen, Chapter 20, pp. 421‐430.) 3 .(15 points) A protein has three sites that can be labeled by chromophores (dyes) by site‐specific chemical modification of appropriate amino acids. The arrangement of these sites is such at 1.5 RAC = RAB = 45 Å. (see figure below) If site A is labeled with a dye (donor), the measured quantum yield (φD) is 0.5. Modification of site A with a donor dye and site B with an acceptor dye results in a measured quantum yield (φD+A) of 0.4. If site A is labeled with donor and both sites B and C are modified with the acceptor dye, predict what will be the measured quantum yield (φD+2A). 4. (10 points) The DNA bases have only very weak fluorescence. To study the unwinding kinetics of double‐ stranded DNA by a helicase, an oligonucleotide is synthesized by substituting one of the adenine with a fluorescent analog, 2‐aminopurine, which can hydrogen‐bond to thymine without distorting the normal DNA structure. When this oligonucleotide hybridizes to a complementary strand, the fluorescence is quenched to half of its original intensity; however, when the oligonucleotide is totally unwound by the helicase, the maximum fluorescence intensity is restored. If 75% of the maximum fluorescence intensity remains after incubating the double‐ stranded oligonucleotide with the helicase for 15 s, what is the rate constant for the unwinding of the oligonucleotide by the helicase? Explain any assumption you make. Short version: When the fluorescent oligonucleotide hybridized to the complementary strand is totally unwound by the helicase, the fluorescence is Fo. When the fluorescent oligonucleotide is hybridized to the complementary strand, F = 0.5 Fo. Now, 75% of the maximum fluorescence remains after incubating the double-stranded oligonucleotide by the helicase for 15 s. So, 50% of the double-stranded oligonucleotide has been unwound by the helicase in 15 s. Assuming the unwounding is a first-order process, t½ = ln2 / k = 15 s Therefore, rate constant k = (ln2) /15 s-1. Longer version: 5. (25 points) (a) Consider a spherical protein with radius r. Set up an integral that could determine the root‐ mean‐square (rms) distance between two randomly selected points within this spherical volume. Evaluate this integral in terms of r using computing programs such as Mathematica or Matlab. Show that this rms distance is [(6/5)]½ r. (20 points) From cosine law, Rab2 = ra2 + rb2 – 2ra rb cos θ Now ra can go from 0 to r, the radius of the protein. Also rb can go from 0 to r. To find the average Rab, we must integrate the above expression over the sphere and normalize the result. Rab 2 = = ∫∫∫ ∫ r r π 2π 0 r 0 0 0 (ra +rb − 2ra rb cosθ )ra rb sin2 θ dϕdθdra drb 2 2 2 2 r ∫∫∫ ∫ π 0 0 0 2π 0 ra rb sin2 θ dϕdθdra drb 2 2 2π r 15 6 2 =r π 2r 6 9 5 28 Rab (root − mean − square) = 6 r 5 (b) Suppose that this spherical protein has a radius 17 Å, and it contains one tyrosine residue and one tryptophan residue, which act as a fluorescence donor‐acceptor pair. Using your result from (a), calculate the FRET efficiency from the tyrosine to the tryptophan if the two residues are separated by the root‐mean‐square distance as calculated. What does the magnitude of this efficiency tell you about the suitability of using this D‐A pair to study this protein? (5 points) FRET efficiency = E = { Ro6 /(R6 + Ro6)} Ro for tyrosine-tryptophan donor acceptor pair ~ 9 Å From (a), the root-mean-square distance is (6/5)½ r or (6/5)½ 17 Å. Substituting these numbers into E = { Ro6 /(R6 + Ro6)}, we obtain, E = {(9 Å)6 /( [(6/5)½ 17 Å]6 + (9 Å)6)} = 0.013 6. (10 points) When CD spectroscopy is performed on three solutions, each containing 1.0 × 10‐4 M of DNA, 1.0 × 10‐4 M of RNA, and a mixture of the two with unknown concentrations, the following data are obtained. All spectra are taken in a cell of path length 1 cm. λ (nm) DNA RNA Mixture 300 0.10 ‐0.50 ‐0.25 280 3.00 1.60 1.16 260 0.00 6.00 3.12 240 ‐2.20 0.00 ‐0.24 AL‐AR (×10‐4) (a) What are the concentrations of DNA and RNA in the mixture? (6 points) For pure DNA and RNA solutions, AL - AR = (εL – εR) c ℓ = (εL – εR) c For a mixture, (AL - AR)mixture = (εL – εR)DNA cDNA + (εL – εR)RNA cRNA At 260 nm, (εL – εR)DNA = 0, Therefore, (AL - AR)mixture = (εL – εR)RNA cRNA = 6.00 x cRNA = 3.12 x 10-4 Or cRNA = 3.12/6.00 x 10-4 M = 0.52 x 10-4 M (εL – εR)RNA = 0, At 240 nm, Therefore, (AL - AR)mixture = (εL – εR)DNA cDNA = -2.20 x cDNA = -0.24 x 10-4 Or (b) If the mixture is hydrolyzed into its component nucleotides, will each of the following increase or decrease? Briefly explain why. (4 points) (i) Circular dichroism at 260 nm; cDNA = 0.24/2.20 x 10-4 M = 0.11 x 10-4 M CD at 260 nm will decrease. (ii). Absorbance at 260 nm. Absorbance will increase. This problem has been postponed to Problem Set # 3 (Reason: I haven’t covered CD yet!) 7. (20 points) (a) Calculate the exciton splitting for two interacting π → π* transition dipoles in a rigid dipeptide (see figure below). Assume a transition dipole of 5.1 debyes or 2 a.u. (1 a.u. = 2.542 debyes) and a distance between the dipoles of 2.1 Å or 4 a.u. (1 a.u. = 0.05292 nm). The two transition dipoles are parallel, and the angle between the transition dipole and the inter‐dipole distance vector is 120° in each case. Express the result in a.u. and in cm‐1 (1 a.u. = 27.3 eV and l eV = 8000 cm‐1). (15 points) (b) What is the distribution of intensity among the two possible transitions in the absorption band? (5 points) 8. (15 points) The folding of a protein from the unstructured or random coil to the native state is often followed in real time by time‐resolved circular dichroism (CD) and tryptophan fluorescence spectroscopy. (a) Indicate what property of the protein structure is being monitored by the two spectroscopic methods. Briefly describe how the two methods provide information about the structure of the protein. (10 points) (b) The two methods, unfortunately, are sensitive to different stages of the folding of the polypeptide. The CD method seems to be particularly sensitive to the early stages of the folding process, whereas tryptophan fluorescence is primarily sensitive to the late phases. What could be the explanation for this difference in behavior between the two methods? (5 points). This problem has also been postponed to Problem Set # 3 (Reason: I haven’t covered CD yet!) ____________________________________________________________________________________ The end ...
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This note was uploaded on 10/27/2010 for the course BI 110 taught by Professor Richards,j during the Winter '08 term at Caltech.

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